Problem: The sequence $(x_n)$ is defined by $x_1 = 115$ and $x_k = x_{k - 1}^2 + x_{k - 1}$ for all $k \ge 2.$  Compute
\[\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + \frac{1}{x_3 + 1} + \dotsb.\]
Solution: Consider the term $\frac{1}{x_{k - 1} + 1}.$  We can multiply the numerator and denominator by $x_{k - 1},$ to get
\[\frac{x_{k - 1}}{x_{k - 1}^2 + x_{k - 1}} = \frac{x_{k - 1}}{x_k}.\]To get the sum to telescope, we can multiply the numerator and denominator again by $x_{k - 1}$:
\[\frac{x_{k - 1}^2}{x_{k - 1} x_k} = \frac{x_k - x_{k - 1}}{x_{k - 1} x_k} = \frac{1}{x_{k - 1}} - \frac{1}{x_k}.\]Hence,
\begin{align*}
\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + \frac{1}{x_3 + 1} + \dotsb &= \left( \frac{1}{x_1} - \frac{1}{x_2} \right) + \left( \frac{1}{x_2} - \frac{1}{x_3} \right) + \left( \frac{1}{x_3} - \frac{1}{x_4} \right) + \dotsb \\
&= \frac{1}{x_1} = \boxed{\frac{1}{115}}.
\end{align*}